Ta có:
1 1.2 + 1 3.4 + 1 5.6 + ... + 1 49.50
= 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ ⋯ + 1 49 − 1 50 = 1 + 1 3 + 1 5 + ⋯ ⋯ + 1 49 − 1 2 + 1 4 + 1 6 + ⋯ ⋯ + 1 50 = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ ⋯ + 1 50 − 2. 1 2 + 1 4 + 1 6 + ⋯ ⋯ + 1 50 = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ ⋯ + 1 50 − 1 + 1 2 + 1 3 + ⋯ ⋯ + 1 25 = 1 26 + 1 27 + 1 28 + ... + 1 50 ( d p c m )
Chứng Minh Rằng: 1/1.2+1/3.4/1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
Chứng minh rằng:
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
GIÚP MÌNH VS <33333
Chứng minh rằng:\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh rằng:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
chung minh rang 1/1.2+1/3.4+1/5.6+....+1/49.50=1/26+1/27+1/28+....1/50
Chứng minh rằng l
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
chứng minh rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)\(\frac{1}{50}\)
chứng minh rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh 1/1.2 + 1/3.4 +1/5.6 +...... + 1/49.50 =1/26 + 1/27 + ... +1/50
Chứng minh rằng: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
